Question

Suppose the number of dollars spent per week on groceries is normally distributed. If the population standard deviation is 7 dollars, what minimum sample size is needed to be 90% confident that the sample mean is within 3 dollars of the true population mean?

Answer 1

Answer: 15

Step-by-step explanation:

Given : Level of confidence = 0.90

Significance level : $\alpha=1-0.90=0.10$

Critical value : $z_{\alpha/2}=1.645$

Margin of error : $E=\text{ 3 dollars}$

Standard deviation: $\sigma=\text{ 7 dollars}$

The formula to find the sample size : $n=(\dfrac{\sigma\times z_{\alpha/2}}{E})^2$

$\Rightarrow n=(\dfrac{7\times(1.645)}{3})^2=14.7328027778\approx15$

Hence, the minimum sample size needed= 15.